46 MOMENT OF ROTATION PARALLEL FORCES. [CHAP. V 



then draw the force and equilibrium polygon, considering each 

 portion to act at its centre of gravity, and so obtain an equili- 

 brium polygon composed of three lines only. These lines will 

 le tangents to the equilibrium curve. (Art. 76.) We thus have 

 three points of the curve, and its direction at these points. 

 In this manner we may determine as many points as may be 

 necessary, without having the sides of the polygon so short or 

 so numerous as to give rise to inaccuracy. 



43. The above will appear more plainly by consideration 

 of a 



BEAM UNIFORMLY LOADED. 



The curve of load distribution becomes in this case a straight 

 line. The load area is then a rectangle, and hence the load per 

 unit of length is constant. Let us now divide this load area 

 [Fig. 30, PI. 8], into four equal parts, and considering each por- 

 tion as acting at its centre of gravity, assume a scale of force, 

 and draw the force polygon. Since in this case the reactions at 

 the supports must be equal, we take the pole C, in a perpendi- 

 cular to the force polygon at the middle point. This causes the 

 closing line of the equilibrium polygon to be parallel to the 

 beam itself, \vhich is often convenient. \Ve now draw C 0, C 1, 

 etc., and then form the polygon a c e g h. The lines a, a c, 

 c e, etc., of this polygon, are tangent to the moment curve at 

 the points , d,f\ and A, where the lines of division prolonged 

 meet the sides. The curve can now be easily constructed, as 

 will appear from the next Art. 



Moment Curve a Parabola. Suppose we had divided the 

 load area into only two parts, of the length x and I x [Fig. 

 30, PL 8]. Then the moment polygon would be o a k A, and 

 the horizontal projection of the tangent line a k would be \ x 

 + % (l-x) = %l. 



That is, the horizontal projection of any tangent line a Jc to 

 the moment curve limited by o C and C A, is constant. But 

 this is a property of the parabola. The moment curve for a 

 uniform, load is therefore a parabola, symmetrical with respect 

 to the vertical through the centre of the beam. 



If, then, we divide o C and C h into equal parts, and join cor- 

 responding divisions above and below, we can construct any 

 number of tangents in any position. 



[NOTE. We maj prove analytically that the moment curve is a parabola, 



